How do you test the alternating series $Sigma (-1)^(n+1)(1-1/n)$ from n is $[1,oo)$ for convergence?

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Answer 1 · 2017-11-07T21:23:10

See below.

$sum_(k=1)^oo(-1)^k a_k = sum_(k=1)^ooa_(2k-1)-a_(2k) = sum_(k=1)^oo b_k$

but

$b_k = 1-1/(2k-1)-(1-1/(2k)) = -1/((2k)(2k-1))$

so we conclude that $sum_(k=1)^oo b_k$ converges

NOTE:

$sum_(k=1)^oo 1/(2k(2k-1)) le sum_(k=1)^oo 1/(2k)^2 = 1/4 pi^2/6$

How do you test the alternating series Sigma (-1)^(n+1)(1-1/n)Sigma (-1)^(n+1)(1-1/n) from n is [1,oo)[1,oo) for convergence?